C++编程 hdu 1065（贪心题目）

Wooden Sticks

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 3

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Problem Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 

(a) The setup time for the first wooden stick is 1 minute. 
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup. 

You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

Output

The output should contain the minimum setup time in minutes, one per line.

Sample Input

3 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1

Sample Output

213

Source

#include <iostream>using namespace std;int n;struct wooden{    int l;    int w;    int no;};wooden wood[5001];void init(){    int i;    for(i = 0; i < 5001; i ++)    {        wood[i].l = 0;        wood[i].w = 0;        wood[i].no = -1;    }}bool compare(int i, int j){    if(wood[i].l >= wood[j].l && wood[i].w >= wood[j].w)        return true;    else if(wood[i].l <= wood[j].l && wood[i].w <= wood[j].w)        return true;    else        return false;}void check(){    int i;    for(i = 0; i < n; i++)    {        cout << "(" << wood[i].l <<","<< wood[i].w <<"," << wood[i].no <<") ";    }}//int doit()//{//    int i;//    int m = 1;//    for(i = 0; i < n - 1; i++)//    {//        if(compare(i, i + 1))//        {//            m ++;//        }//    }//    return m;//}int main(){    int t, cur, w;    int i, j;    cin >> t;    while(t--)    {        cin >> n;        init();        for(i = 0; i < n; i++)            cin >> wood[i].l >> wood[i].w;        for(i = 0; i < n; i ++)        {            for(j = 0; j < n - i - 1; j++)            {                if(wood[j].l > wood[j+1].l)                {                    swap(wood[j], wood[j+1]);                }            }        }//        for(i = 0; i < n; i ++)//        {//            for(j = 0; j < n - i - 1; j++)//            {//                if(wood[j].w > wood[j+1].w)//                    swap(wood[j], wood[j+1]);//            }////        }//        }        cur = 0;        for(i = 0; i < n; i++)        {            if(wood[i].no == -1)            {                cur ++;                w = wood[i].w;                wood[i].no = cur;                for(j = i; j < n; j++)                {                    if(wood[j].no == -1 && w <= wood[j].w)                    {                        wood[j].no = cur;                        w = wood[j].w;                    }                }            }//            check();//                cout << endl;        }        cout << cur << endl;    }    return 0;}

大水题。。但是做的时候出了问题 = =。一开始的想法是根据l排一次序，再根据w排一次序。。这是不行的，原因如下。。

可能出现这种情况（7，8），（8, 7），（9,10），（10,9），（11，11）

类似于如上情况就行不通了。。doit（）注释掉的就是之前错的算法= =。。看别人说的分集合恍然大悟。。遂写成。。唉，这下惨了。。水题做成这个样子。。